What is the capacitance of a spoon?

Gary Godfrey beat me to the spherical cow joke. For a slightly more sophisticated theorist's answer, let's consider an ellipsoidal cow instead. According to the Digital Library of Mathematical Functions, the inverse of the capacitance of an conducting ellipsoid with semi-major axes $a$, $b$, and $c$ is $frac{1}{C} = R_F(a^2, b^2, c^2) / (4 pi epsilon_0)$, where $$ R_F(a^2, b^2, c^2) = frac{1}{2} int_0^infty frac{dt}{sqrt{(t+a^2)(t+b^2)(t+c^2)}}. $$ (Note that the formula given in the above link is in CGS units; I think I have correctly converted it to MKS units, but let me know if this needs correction.) This integral doesn't have a closed-form expression for arbitrary $a$, $b$, and $c$; but for $a = b < c$, it can be performed: $$ R_F(a^2, a^2, c^2) = frac{cosh^{-1} (c/a)}{sqrt{c^2 - a^2}}. $$ This implies that $$ C = 4 pi epsilon_0 frac{sqrt{c^2 - a^2}}{cosh^{-1} (c/a)}. $$ If we approximate a spoon as an ellipsoid of length 20 cm and diameter 2 cm, we have $c = 10$ cm and $a = 1$ cm, and we obtain $ C approx 3.7$ pF.

As another example, if we approximate a human body as an ellipsoid with $c = 80$ cm and $a = 20$ cm, we obtain $C approx 42$ pF. We can see that this is within an order of magnitude of estimates found elsewhere.

For an object that is better approximated as an oblate ellipsoid, with $a = b > c$, the integral is slightly different, and the capicatnce turns out to be: $$ C = 4 pi epsilon_0 frac{sqrt{a^2 - c^2}}{cos^{-1} (c/a)}. $$ If a frying pan has a radius of $c approx 15$ cm, and a thickness of about 4 cm (so $a approx 2 cm$), then $C approx 11.5 pF$. Still smaller than that of a human body. $$

Finally, note that in both cases, for a given ratio of $c/a$, the capacitance of a body scales linearly with its size. This is a general property that can (I think) be proven rigorously for bodies of arbitary shape via arguments based on the properties of Laplace's equation.